I am trying to solve the following exercise:
Let $(X,d)$ be a metric space that has the property that for any $x\in X$ and $r>0$, the closed ball
$$\bar{B}(x,r):=\{y\in X:d(x,y)\leq r\}$$
is compact. Show that $X$ is complete.
I think I have a proof, but I am only using that the unit closed balls $\bar{B}(x,1)$ are compact. Maybe I am missing something?
Attempt: Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $(X,d)$. Then, $\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(x_n,x_N)<1$. Thus, $\forall n\geq N$, $x_n\in\bar{B}(x_N,1)$ which is compact by assumption, so $(x_n)_{n=N}^\infty$ is a sequence in this compact set and thus has a convergent subsequence. Hence, $(x_n)_{n=1}^\infty$ is a Cauchy sequence that has a convergent subsequence, so it converges in $(X,d)$. Therefore, $(X,d)$ is complete.
Best Answer
Your proof is right.
As a justification that it is not strange that you needed a little less than the hypothesis, let me show you what happens when your metric space is a normed space: since any two balls are homeomorphic, if one closed ball is compact than all are. Indeed, $$ \bar B(0,1)\simeq \bar B(y,r) $$ via the continuous function $x\mapsto r(x+y)$.